steady-state method to calculate thermal and electrical conductivity

1. THERMAL AND ELECTRICAL
CONDUCTIVIT Y OF METAL
S O U R AV G H O S H
1 5 2 0 8
PHY 405
CONDENSED MATTER PHYSICS LAB

2. AIM OF THE EXPERIMENT
EXPERIMENTAL SETUP
• Determine the thermal conductivity of the metal rod(Copper/Aluminium)
• Determine the electrical conductivity of Copper and Aluminium rod.
• To test theWiedemann-Franz law and find out the Lorentz number
Source- indosaw manual

3. PROCEDURE
❖Thermal conductivity measurement-
1.Measurement of heat capacity of lower calorimeter
𝑐 = 𝑐 𝑤. 𝑚 𝑤.
𝑇 𝑊−𝑇 𝑚
𝑇 𝑚−𝑇 𝑅
Where, 𝑐 𝑤=Specific heat capacity of water, 𝑚 𝑤=Mass of water, 𝑇 𝑊=Temperature of hot water,
𝑇 𝑚=Temperature of mixing, 𝑇𝑅=Room temperature
2. Determination of influence of the surroundings-
The addition of heat from the surroundings is calculated from the temperature increase in the lower
calorimeter (for that we have to fill the lower calorimeter with ice and check the temperature of the lower
calorimeter in 1 min interval till the temperature rises to 12 degree Centigrade.
𝑄 = 𝐶𝑤 ⋅ 𝑚 𝑤 + 𝐶 ⋅ 𝑇 − 𝑇0
𝑇0 =Temperature at time t=0
Slope of the graph Q vs t gives us (dQ/dt)surr
3. Determination of the heat flow through metal rod and calculation of thermal conductivity-
ⅆ𝑄 𝑟𝑜𝑑
ⅆ𝑡
= −𝜆𝐴
Δ𝑇
Δ𝑥
For that we have to maintain almost constant temperature gradient between the upper calorimeter and
the lower calorimeter at a constant temperature and note the differential temperature and the temperature
of the lower calorimeter .
Now slope of Q-t graph will give us (dQ/dt)total and (dQ/dt)rod=(dQ/dt)total-(dQ/dt)surr

4. PROCEDURE…CONTINUED
❖Electrical Conductivity Measurement
– Connected the experiment setup according to the circuit
– Slowly increased the current value in power supply and noted the values of current and voltage
– From theV-I graph slope we will get R
– As we know, 𝜎 =
𝐿
𝐴𝑅
where , 𝜎= electrical conductivity, L=length of the rod,A=cross section
area of hot end

5. THEORY
• Thermal conductivity of a material is a measure of its ability to conduct heat.
• The quantity of heat transported with time dt is a function of the cross-sectional area A and
the temperature gradient ΔT/Δx perpendicular to the surface.
𝑑𝑄
𝑑𝑡
= −𝜆𝐴
Δ𝑇
Δ𝑥
….. |a| 𝜆=thermal conductivity of the material
• The temperature distribution in a body is generally a function of location and time and is in
accordance with the Boltzmann transport equation
𝑑𝑇
𝑑𝑡
=
𝜆
𝜌𝑐
⋅
𝑑2 𝑇
𝑑𝑥2 ….|b|
• After some time, a steady state is reached when the two ends are maintained at constant
temperature.
𝑑𝑇
𝑑𝑡
= 0
• In order to calculate the heat energy transported by the metal rod according to(a), the
ambient contribution from the surroundings must be subtracted
𝑑𝑄 𝑟𝑜𝑑
𝑑𝑡
=
𝑑𝑄𝑡𝑜𝑡
𝑑𝑡
−
𝑑𝑄 𝑠
𝑑𝑡

6. WIEDEMANN-FRANZ LAW..
RELATION BETWEEN ELECTRICAL CONDUCTIVITY AND THERMAL CONDUCTIVIT Y
• At room temperature the conduction electrons in metal have much greater mean free path
compared to phonons and therefore heat conduction in metals is primarily due to the
electrons.The correlation between the thermal conductivity λ and electrical conductivity σ is
given by theWiedemann Franz law-
𝜆
𝜎
= 𝐿 ⋅ 𝑇
where 𝜆= thermal conductivity , 𝜎= electrical conductivity , L=Lorentz number
theoretical value of L,
𝐿 =
𝜋2
3
⋅
𝑘2
ⅇ2 = 2 ⋅ 4 × 10−8 𝑤𝛺
𝑘2

7. RESULTS…
COPPER

8. RESULTS…
ALUMINIUM

9. RESULTS
❖ Copper
➢ Heat Capacity of calorimeter = 58.17 Cal/K
➢Thermal conductivity, λ = 200.34W/m-K
■ Literature value = 385W/m-K
➢ Electrical conductivity, σ = 5.185*107 S/m
■ Literature value = 5.96*107 S/m
➢ Lorenz number, L = 1.305*10-8W-Ω/K2
■ Literature value = 2.44*10-8W-Ω/K2

10. RESULTS
❖ Aluminium
➢Thermal Conductivity, λ = 89.46W/K-m
■ Literature value = 205W/K-m
➢ Electrical Conductivity, σ = 3.352*107 S/m
■ Literature value = 3.77*107 S/m
➢ Lorenz Number, L = 0.9*10-8W-Ω/K2
■ Literature value = 2.44*10-8W-Ω/K2

11. || T H A N K Y O U ||